Theorem(Long and Niblo): If $M$ is a 3-manifold and $S$ is an incompressible component of $\partial M$ then $\pi_1 S$ is separable in $\pi_1 M$ (pick a base point in $S$ to make sense of this).

This means that for any $\gamma\in \pi_1M\smallsetminus \pi_1 S$ there is a homomorphism $f$ from $\pi_1M$ to a finite group such that $f(\gamma)\notin f(\pi_1(S))$.

Corollary: If $|\pi_1M:\pi_1S|=\infty$ then $M$ had finite-sheeted coverings such that the preimage of $S$ has as many components as you like.

Proof: Choose a homomorphism $f$ such that $f(\pi_1S)$ has large index in $f(\pi_1M)$. Now your covering corresponds to $\ker f$. QED

So examples exists as finite covers of the example you already have!

Theorem(Long and Niblo): If $M$ is a 3-manifold and $S$ is an incompressible component of $\partial M$ then $\pi_1 S$ is separable in $\pi_1 M$ (pick a base point in $S$ to make sense of this).

This means that for any $\gamma\in \pi_1M\smallsetminus \pi_1 S$ there is a homomorphism $f$ from $\pi_1M$ to a finite group such that $f(\gamma)\notin f(\pi_1(S))$.

Corollary: If $|\pi_1M:\pi_1S|=\infty$ then $M$ had finite-sheeted coverings such that the preimage of $S$ has as many components as you like.

Proof: Choose a homomorphism $f$ such that $f(\pi_1S)$ has large index in $f(\pi_1M)$. Now your covering corresponds to $\ker f$. QED

So examples exists as finite covers of the example you already have!